ME3120 · Dynamic Modelling

Theme 2 · Newton's method

Checking a model

You have built several models now. Before you trust one — or simulate it — run three fast checks that catch most mistakes: units, limits, and equilibrium.

Source: course notes — the engineer's three-point model sanity check.

Before you start

What you need first

The mass–spring–damper $m\ddot x + c\dot x + kx = F$, and the SI units of each quantity.
The pendulum $\ddot\theta + \tfrac{g}{L}\sin\theta = 0$ and $\omega_n=\sqrt{g/L}$.

What you'll be able to do

Check 1 — units: confirm every term matches.
Check 2 — limits: test the extremes for sense.
Check 3 — equilibrium: find where the system rests, and whether it is stable.

Why check a model?

A model is just algebra you wrote — it can hide a dropped sign, a missing factor, or a wrong term. If you simulate it without checking, a wrong model gives confident, wrong answers.

Three quick checks catch most errors in seconds, before you trust the model: (1) do the units match, (2) does it behave at the limits, and (3) where does it rest? A good model passes all three.

Check 1

Do the units match?

Every term you add together must carry the same units — you cannot add newtons to metres. Run the mass–spring–damper, term by term:

Term	Units, multiplied out	Result
$m\ddot x$	$\text{kg}\times\text{m/s}^2$	N
$c\dot x$	$(\text{N·s/m})\times(\text{m/s})$	N
$kx$	$(\text{N/m})\times\text{m}$	N
$F$	—	N

All four are newtons → the equation is dimensionally consistent. If one term came out in different units, you would know a factor is missing or a term is wrong.

Angles are unit-free. A radian is a length over a length, so it carries no units. That is why the pendulum's $\ddot\theta$ ($\text{rad/s}^2=1/\text{s}^2$) and $\tfrac{g}{L}\sin\theta$ ($\tfrac{\text{m/s}^2}{\text{m}}=1/\text{s}^2$) also match.

Check 2

Does it behave at the limits?

Push a parameter to an extreme and ask: does the prediction still make sense? Use the pendulum frequency $\omega_n=\sqrt{g/L}$:

Limit	$\omega_n$	Sense?
$L\to\infty$ (very long)	$\to 0$	swings very slowly ✓
$L\to 0$ (very short)	$\to\infty$	swings very fast ✓
$g\to 0$ (in space)	$\to 0$	no gravity, never swings back ✓

All three match intuition → the formula earns trust. A model that predicted a longer pendulum swinging faster would be warning you that something is wrong.

$\omega_n=\sqrt{g/L}$: as $L$ grows the frequency falls toward zero; as $L$ shrinks it shoots up — exactly the sensible behaviour.

Check 3

Where does it rest?

An equilibrium is a state where nothing changes — so every velocity and acceleration is zero. Put $\dot\theta=\ddot\theta=0$ into the pendulum equation and solve:

1Set $\ddot\theta=0$ in $\ddot\theta+\tfrac{g}{L}\sin\theta=0$:

$$\frac{g}{L}\sin\theta = 0 \;\Rightarrow\; \sin\theta = 0$$

2Solve $\sin\theta=0$ over one turn:

$$\theta = 0 \quad\text{or}\quad \theta = \pi$$

Two rest positions: hanging straight down ($\theta=0$) and balanced straight up ($\theta=\pi$). Both are real — but they do not feel the same.

Stable or unstable?

Give each a tiny nudge and watch the torque:

Down ($\theta=0$) — stable. A nudge makes gravity pull it back; it returns and swings about the bottom.
Up ($\theta=\pi$) — unstable. A nudge makes gravity pull it further away; it falls. Balancing it there needs perfect aim.

Stable = the system comes back after a small disturbance. Unstable = it runs away. The equilibrium check tells you where it can rest; the nudge test tells you whether it will stay.

Down ($\theta=0$): a nudge is pushed back — stable. Up ($\theta=\pi$): a nudge is pushed away — unstable.

📐 Worked example

Check a hanging-spring model

A mass $m = 2\ \text{kg}$ hangs from a vertical spring of stiffness $k = 400\ \text{N/m}$. Measuring $x$ downward from the spring's natural length, a proposed model is $m\ddot x = mg - kx$. Take $g = 9.81\ \text{m/s}^2$ and run the three checks.

A $2\ \text{kg}$ mass on a vertical spring; $x$ is measured downward from the natural length.

1 · UnitsCheck each term:

$$\underbrace{m\ddot x}_{\text{kg·m/s}^2=\text{N}} \;=\; \underbrace{mg}_{\text{N}} \;-\; \underbrace{kx}_{(\text{N/m})\text{m}=\text{N}}$$

2 · EquilibriumSet $\ddot x=0$ and solve for the rest position $x_{\text{eq}}$:

$$0 = mg - kx_{\text{eq}} \;\Rightarrow\; x_{\text{eq}} = \frac{mg}{k} = \frac{(2)(9.81)}{400} = 0.049\ \text{m}$$

3 · LimitsTest $x_{\text{eq}} = mg/k$ at the extremes:

$$k\to\infty \Rightarrow x_{\text{eq}}\to 0,\qquad m\to 0 \Rightarrow x_{\text{eq}}\to 0$$

All three pass: units consistent, the mass hangs $49\ \text{mm}$ below natural length (heavier mass → more stretch, stiffer spring → less), and the extremes are sensible. The model is trustworthy.

✏️ Try it yourself

Run the three checks on the horizontal mass–spring $m\ddot x + kx = 0$, with $m = 0.5\ \text{kg}$ and $k = 200\ \text{N/m}$.

Units. $m\ddot x$: $\text{kg·m/s}^2=\text{N}$; $kx$: $(\text{N/m})\,\text{m}=\text{N}$. Match ✓ Equilibrium. $\ddot x=0 \Rightarrow kx=0 \Rightarrow x=0$ — rests at the spring's natural length. Limits. $\omega_n=\sqrt{k/m}=\sqrt{200/0.5}=20\ \text{rad/s}$; as $k\to 0$, $\omega_n\to 0$ (no spring, no oscillation) ✓

Common mistakes to avoid

Mistake	Fix
Skipping the checks and simulating straight away	A wrong model gives confident wrong answers — spend 30 seconds checking first.
Forgetting that radians are unit-free	$\sin\theta$ and angles are dimensionless; only the $g/L$ part carries units.
Looking for only one equilibrium	$\sin\theta=0$ has more than one solution ($\theta=0$ and $\theta=\pi$) — list them all.
Assuming every equilibrium is stable	Always run the nudge test; the upright pendulum rests there but will not stay.
Testing a limit and ignoring a wrong result	If a limit looks absurd, trust the check — go back and find the modelling error.

Recap — the whole topic on one screen

Check	How	What it catches
1 · Units	Every term in the same units	missing factors, wrong terms
2 · Limits	Push a parameter to an extreme	wrong dependence (e.g. sign, power)
3 · Equilibrium	Set velocities & accelerations to $0$, solve	wrong rest state; reveals stability

These three checks close out Theme 2 — Newton's method. You can now build a single-DOF model and prove it sane. Next we learn a second way to build models — from energy.

Next theme

Kinetic & potential energy

Newton's method means drawing forces and torques. For complicated machines that gets messy. Theme 3 builds the same models a different way — by bookkeeping energy. First we meet the two kinds: kinetic (motion) and potential (position).

→ Kinetic & potential energy

Term	Units, multiplied out	Result
\(m\ddot x\)	\(\text{kg}\times\text{m/s}^2\)	N
\(c\dot x\)	\((\text{N·s/m})\times(\text{m/s})\)	N
\(kx\)	\((\text{N/m})\times\text{m}\)	N
\(F\)	—	N

Limit	\(\omega_n\)	Sense?
\(L\to\infty\) (very long)	\(\to 0\)	swings very slowly ✓
\(L\to 0\) (very short)	\(\to\infty\)	swings very fast ✓
\(g\to 0\) (in space)	\(\to 0\)	no gravity, never swings back ✓